Ask question

Ask question

All categories

3 years ago

8

Owen settled on a price of $7,870 for a new car. The dealer had to add 4.5% sales tax to this price, but allowed a $1,200 trade-

in for Owen's old car. Although not a universal practice, please add the sales tax to the price of the new car first and then deduct the trade- in value. If the dealer required a 15% down payment, calculate the amount of the purchase price that Owen financed.

1 answer:

kakasveta [241]3 years ago

4 0

Multiply the numbers by the percent but first convert the percent into a decimal then after u multiply add that number to the tottal

You might be interested in

[Calculus, Derivatives] Can a function be differentiable at certain points even if the entire function is not differentiable?

koban [17]

A function is differentiable if you can find the derivative at every point in its domain. In the case of f(x) = |x+2|, the function wouldn't be considered differentiable unless you specified a certain sub-interval such as (5,9) that doesn't include x = -2. Without clarifying the interval, the entire function overall is not differentiable even if there's only one point at issue here (because again we look at the entire domain). Though to be fair, you could easily say "the function f(x) = |x+2| is differentiable everywhere but x = -2" and would be correct. So it just depends on your wording really.

5 0

4 years ago

Z^4-5(1+2i)z^2+24-10i=0

mixer [17]

Using the quadratic formula, we solve for $z^2$ .

$z^4 - 5(1+2i) z^2 + 24 - 10i = 0 \implies z^2 = \dfrac{5+10i \pm \sqrt{-171+140i}}2$

Taking square roots on both sides, we end up with

$z = \pm \sqrt{\dfrac{5+10i \pm \sqrt{-171+140i}}2}$

Compute the square roots of -171 + 140i.

$|-171+140i| = \sqrt{(-171)^2 + 140^2} = 221$

$\arg(-171+140i) = \pi - \tan^{-1}\left(\dfrac{140}{171}\right)$

By de Moivre's theorem,

$\sqrt{-171 + 140i} = \sqrt{221} \exp\left(i \left(\dfrac\pi2 - \dfrac12 \tan^{-1}\left(\dfrac{140}{171}\right)\right)\right) \\\\ ~~~~~~~~~~~~~~~~~~~~= \sqrt{221} i \left(\dfrac{14}{\sqrt{221}} + \dfrac5{\sqrt{221}}i\right) \\\\ ~~~~~~~~~~~~~~~~~~~~= 5+14i$

and the other root is its negative, -5 - 14i. We use the fact that (140, 171, 221) is a Pythagorean triple to quickly find

$t = \tan^{-1}\left(\dfrac{140}{171}\right) \implies \cos(t) = \dfrac{171}{221}$

as well as the fact that

$0$

$\sin\left(\dfrac t2\right) = \sqrt{\dfrac{1-\cos(t)}2} = \dfrac5{\sqrt{221}}$

(whose signs are positive because of the domain of $\frac t2$ ).

This leaves us with

$z = \pm \sqrt{\dfrac{5+10i \pm (5 + 14i)}2} \implies z = \pm \sqrt{5 + 12i} \text{ or } z = \pm \sqrt{-2i}$

Compute the square roots of 5 + 12i.

$|5 + 12i| = \sqrt{5^2 + 12^2} = 13$

$\arg(5+12i) = \tan^{-1}\left(\dfrac{12}5\right)$

By de Moivre,

$\sqrt{5 + 12i} = \sqrt{13} \exp\left(i \dfrac12 \tan^{-1}\left(\dfrac{12}5\right)\right) \\\\ ~~~~~~~~~~~~~= \sqrt{13} \left(\dfrac3{\sqrt{13}} + \dfrac2{\sqrt{13}}i\right) \\\\ ~~~~~~~~~~~~~= 3+2i$

and its negative, -3 - 2i. We use similar reasoning as before:

$t = \tan^{-1}\left(\dfrac{12}5\right) \implies \cos(t) = \dfrac5{13}$

$1 < \tan(t) < \infty \implies \dfrac\pi4 < t < \dfrac\pi2 \implies \dfrac\pi8 < \dfrac t2 < \dfrac\pi4$

$\cos\left(\dfrac t2\right) = \dfrac3{\sqrt{13}}$

$\sin\left(\dfrac t2\right) = \dfrac2{\sqrt{13}}$

Lastly, compute the roots of -2i.

$|-2i| = 2$

$\arg(-2i) = -\dfrac\pi2$

$\implies \sqrt{-2i} = \sqrt2 \, \exp\left(-i\dfrac\pi4\right) = \sqrt2 \left(\dfrac1{\sqrt2} - \dfrac1{\sqrt2}i\right) = 1 - i$

as well as -1 + i.

So our simplified solutions to the quartic are

$\boxed{z = 3+2i} \text{ or } \boxed{z = -3-2i} \text{ or } \boxed{z = 1-i} \text{ or } \boxed{z = -1+i}$

3 0

1 year ago

PLEASE Help me asaaaaap

Sholpan [36]

C is your answer....

3 0

4 years ago

SUPER EASY PLEASE HELP! 15 POINTS.

Anastasy [175]

Answer:

x=-3

Step-by-step explanation:

6 0

3 years ago

You start at (0, -2). You move down 3 units and left 4 units. Where do you end?

tresset_1 [31]

Answer:

-4,-5?

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers